Why It Couldn’t Be The short story Cold Equations by Tom Godwin takes place on a ship called EDS. The space cruiser is piloted by a man named Barton. He has an order of killing the stowaway who snuck onto the ship because the weight on the EDS is too much for the ship to handle. In the process of hunting down the stowaway, he realizes it was a young innocent girl named Marilyn. Once Barton understands what kind of person Marilyn is, he doesn’t kill her immediately because he knows her reasons were pure. Marilyn only wanted to see her brother, Gerry, again after ten years of being apart and was ignorant to the fact that her life can end with the decision of sneaking onto the ship. Barton begins to feel compassion after being with her and tries to comfort her, but knows what her fate is. He lets Marilyn live long enough to let her speak with Gerry once more before he follows through with the command. After Gerry and Marilyn speak he ejects her out into space. The ending was logical and no other endings would be possible because one the equation that was calibrated delicately, and two Barton could not throw the out the fever serums because that is the main reason for going on the trip to Woden. A theoretical ending of Cold Equations could have been that Barton sacrifices himself for Marilyn, but since she is lighter than him, the fragile calibrated equation would be disrupted due to the change in weight. On EDS everything on ship is accounted for its’ weight in an equation that takes in all the factors of the ship. “The white hand of the tiny gauge had crept up. There was something in the supply closet across the room, some kind of a body that radiated heat.” (Godwin 164) Obviously, when Marilyn snuck on to the ship Barton already knew something extra was on board and he would have to get rid of it. “Any stowaway discovered in an EDS shall be jettisoned immediatelyfollowing discovery. It was law and there could be no appeal.” (Godwin 165) Marilyn’s weight made...

YOU MAY ALSO FIND THESE DOCUMENTS HELPFUL

...portrayal of the setting and atmosphere in The ColdEquations, the reader undoubtedly experiences the lonesome and cold feeling occurring in the story. The authors brilliant use of figurative language and imagery illustrating Bartons occupation produces a character deficient of personality and feeling. This genius utilization of words forces the reader to experience a sense of urgency amidst the bitter conditions within the story. TheColdEquations is a fictional tale consumed with chilling imagery expertly depicted by the author, vividly generating a cold atmosphere lacking personality and emotion.
Goodwins effective delineation of the frontier, the setting of which the story takes place, initiates the lonesome and cold aspect experienced in the tragic tale. The ColdEquations occurs in a futuristic time period, resulting in a scientific feel and a lack of understanding by the reader. The Stardust had gone through the usual procedure, dropping into normal space to launch the EDS with the fever serum, then vanishing again in hyperspace. (Goodwin 3). The vast expanse of space cannot be comprehended by the human mind, resulting in the lonesome feeling previously stated. During the commencement of the story, much of the diction contains technical and scientific language. Science is free from emotion and personality, forming the cold...

...
My Mathematical Equations
By Kathleen Rossi
MAT 222 Intermediate Algebra
Instructor: Mohamed Elseifieen
May 12, 2013
My Mathematical Equations
This paper will show two mathematical problems, the first is “To estimate the size of the bear population on the Keweenaw Peninsula, conservationists captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservationist's estimate of the size of the bear population?” (Dugopolski, 2013, pp. 437, probem 56). The second will be to complete problem 10 on page 444 of Elementary and Intermediate Algebra. Here all steps in solving the problem will be explained step by step.
The first problem is to estimate the size of the bear population located on the Keweenaw Peninsula conservation. In reading over the “Bear Population” method #56 on page 437you will notice we are to assume that the ratio of originally tagged bears to the whole population is equal to the ratio of recaptured bears to the size of the sample.
The ratio of the originally tagged bears to the whole population is 2100
The ration of the recaptured tagged bears to the sample size is 50x
2100=50x Since x is on the right-hand side of the equation, we need to switch the sides so it is on the left-hand side.
50x=2100 This is the proportion set up and ready to solve. I will cross multiply setting the extremes equal to the means.
100(50) = 2x Here 100...

...How Cold Affects The Body
B Y: G E N E S I S P O N C E
How Cold Affects The Body
The body tries to maintain body temperature by
vasoconstriction and shivering.

Shivering is the body’s main involuntary defense against the cold
producing body heat by forcing muscles to contract and relax
rapidly.
Vasoconstriction is the tightening of blood vessels
 Vasoconstriction occurs in the uncovered skin when it is exposed
to cold temperatures.
 The reduced blood flow in the skin conserves body heat but can
lead to discomfort, numbness, loss of dexterity in the hands and
fingers.
How The Body Produces Heat
Normal body temperature is maintained by a
balance of heat production and heat loss
Heat is produced by food metabolism and
muscle activity.

Shivering increases heat production up to 500%
 lack of food limits the body’s ability to
produce heat.
Heat Loss From The Body
Body heat can be lost by four mechanisms.
1. Convection is the loss of heat from the body by air
blowing over the skin or through clothing.
2. Conduction, or direct contact with a colder object.
For example, lying in snow but with the exception of
being immersed in cold water; Heat is then lost 25
to 30 times faster.
3. Evaporation, or conversion of liquid on the skin to
vapor. This accounts to 20% of heat loss through
sweating and respiration.
4. Radiation is the primary method of heat loss,
accounting for about 65% of the...

...Quadratic Equation:
Quadratic equations have many applications in the arts and sciences, business, economics, medicine and engineering. Quadratic Equation is a second-order polynomial equation in a single variable x.
A general quadratic equation is:
ax2 + bx + c = 0,
Where,
x is an unknown variable
a, b, and c are constants (Not equal to zero)
Special Forms:
* x² = n if n &lt; 0, then x has no real value
* x² = n if n &gt; 0, then x = ± n
* ax² + bx = 0 x = 0, x = -b/a
WAYS TO SOLVE QUADRATIC EQUATION
The ways through which quadratic equation can be solved are:
* Factorizing
* Completing the square
* Derivation of the quadratic formula
* Graphing for real roots
Quadratic Formula:
Completing the square can be used to derive a general formula for solving quadratic equations, the quadratic formula. The quadratic formula is in these two forms separately:
Steps to derive the quadratic formula:
All Quadratic Equations have the general form, aX² + bX + c = 0
The steps to derive quadratic formula are as follows:
Quadratic equations and functions are very important in business mathematics. Questions related to quadratic equations and functions cover a wide range of business concepts that includes COST-REVENUE, BREAKEVEN ANALYSIS, SUPPLY/DEMAND &amp; MARKET EQUILIBRIUM....

...﻿Quadratic equation
In elementary algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of the equation, and may be distinguished by calling them, the quadratic coefficient, the linear coefficient and the constant or free term.
Solving the quadratic equation
A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Factoring by inspection
It may be possible to express a quadratic equation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadratic equation is written in the second form, then the "Zero Factor Property" states that the quadratic equation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear equations provides the roots of the quadratic.
Completing the square
The process of completing the square makes use of the algebraic identity...

...There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c
Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c
Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will...

...Summer 2010-3 CLASS NOTES CHAPTER 1
Section 1.1: Linear Equations
Learning Objectives:
1. Solve a linear equation
2. Solve equations that lead to linear equations
3. Solve applied problems involving linear equations
Examples:
1. [pic]
[pic]
3. A total of $51,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is to exceed that in CDs by $3,000, how much will be invested in each type of investment?
4. Shannon, who is paid time-and-a-half for hours worked in excess of 40 hours, had gross weekly wages of $608 for 56 hours worked. What is her regular hourly wage?
Answers: 1. [pic]
2. [pic]
3. $24,000 in CDs, $27,000 in bonds 4. $9.50/hour
Section 1.2: Quadratic Equations
Learning Objectives:
1. Solve a quadratic equation by (a) factoring, (b) completing the square, (c) the
quadratic formula
2. Solve applied problems involving quadratic equations
Examples:
1. Find the real solutions by factoring: [pic]
2. Find the real solutions by using the square root method: [pic]
3. Find the real solutions by completing the square: [pic]
4. Find the real solutions by using the quadratic formula: [pic]
5. A ball is thrown vertically upward from the top of a...

...Balancing Equations
Balancing equations is a fundamental skill in Chemistry. Solving a system of linear equations is a fundamental skill in Algebra. Remarkably, these two field specialties are intrinsically and inherently linked.
2 + O2 ----> H2OA. This is not a difficult task and can easily be accomplished using some basic problem solving skills. In fact, what follows is a chemistry text's explanation of the situation:
Taken from: Chemistry
Wilberham, Staley, Simpson, Matta
Addison Wesley
1. Determine the correct formulas for all the reactants and products in the reaction.
2. Write the formulas for the reactants on the left and the formulas for the products on the right with an arrow in between. If two or more reactants or products are involved, separate their formulas with plus signs.
3. Count the number of atoms of each element in the reactants a products. A polyatomic ion appearing unchanged on both sides of the equation is counted as a single unit.
4. Balance the elements one at a time by using coefficients. A is a small whole number that appears in front of a formula a an equation. When no coefficient is written, it is assumed to be 1. It is best to begin with an element other than hydrogen or oxygen. These two elements often occur more than twice in an equation.
5. Check each atom or polyatomic ion to be sure that the equation is balanced....